= Solution
Use the <Minkowski metric> with signature $(+---)$ and write $U=(U^0,\mathbf u)$, $V=(V^0,\mathbf v)$. Future-pointing timelike means $U^0>\|\mathbf u\|$ and $V^0>\|\mathbf v\|$. Thus
$$
U^0+V^0>\|\mathbf u\|+\|\mathbf v\|\geq\|\mathbf u+\mathbf v\|,
$$
so $U+V$ is again future-pointing timelike.
Let $a=\sqrt{g(U,U)}>0$ and $b=\sqrt{g(V,V)}>0$. In a <Lorentz frame> where $U=(a,0,0,0)$, the future-pointing vector $V$ has $V^0\geq b$. Consequently $g(U,V)=aV^0\geq ab$, with equality exactly when $V$ has no spatial component in that frame. Hence
$$
g(U+V,U+V)=a^2+b^2+2g(U,V)\geq(a+b)^2.
$$
Taking positive square roots proves
$$
\boxed{\sqrt{g(U+V,U+V)}\geq\sqrt{g(U,U)}+\sqrt{g(V,V)}.}
$$
\b[Equality holds exactly when $U,V$ are positive scalar multiples of each other.] This is the <reverse triangle inequality for future timelike vectors>, not the Euclidean triangle inequality.
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